# titrate sulphuirc acid with sodium hydroxide

by The Bob
Tags: acid, hydroxide, sodium, sulphuirc, titrate, titration
 P: 1,116 Coursework time has hit me at college and no matter how much I hit the books the answer is not hitting me. It might be simple but I can't see it. Of course, the question is needed. I had to titrate sulphuirc acid with sodium hydroxide. Not hard to write coursework on and I am sure that if I did it without what I am trying to find out (below) then it would be fine but I am not happy just to accept that there should be equal amounts of sulphuric acid and sodium hydroxide to create a neutral solution. I read in these forums that the molar pH of sulphuirc acid is less than that of the pOH of the sodium hydroxide. For my hypothesis I want to be able to say why I think that there needs to be more sodium hydroixde to create a neutral pH overall. For what I need help with is the maths, really. I have been studying for the last 4 hours trying to make sense of the dissonance(?) of acids and bases and how this affects the pH or the moles (not even sure which). I installed a system that was posted up here from the 'few questions' thread (called BATE) and it says that for an equal amount of $$0.01$$ $$mol$$ $$dm^{-3}$$ concentrations of each susbstance I should get an overall pH of 6.17. I am trying to find out how to get to this number and how I can explain it along the way. Even if this wasn't coursework I would have ask the same question (actually, to be honest I don't need it in my coursework it is more for interest but I really want to understand it). Cheers The Bob (2004 ©) P.S. After re-reading it I hope it makes sense. What I want to know is how to work out that the pH after mixing, say, 10cm³ of sodium hydroixde to 10cm³ of sulphuric acid (both concentrated to 0.01M) is going to be 6.17, and therefore how much sodium hydroxide is actually needed. Also I will need to explain why so if you know any websites that will do that, to save you writing it, then I would be grateful.
 Sci Advisor HW Helper P: 1,769 The pH of an equimolar sodium hydroxide and sulfuric acid is 7, both are considered strong acid/base; the product salt of the two is weak with respect acid or base and thus the pH will remain 7. In the case of a different combination of where one of them is a weak acid/base and the other a strong base/acid with equimolar values of both, you'll need to consider that the product salt component of the weak acid/base will actually influence the pH upon the complete neutralization. Yes, the strong base acid/base will react to completion with the weak acid/base (that is completely neutralized)...the sole factor which affects the pH afterwards is the product salt.
 P: 1,116 I have done the titration and the results show that for 10.0cm³ of 0.01M sulphuric acid, approximately 11.0cm³ of 0.01M sodium hydroixde is needed. Are you trying to say that on of the substances is strong and the other is weak because with equal concentrations I can't see that making any difference. I have been looking around and I found these sorts of equations: $$K_w = [H^+][OH^-]$$ $$K_a = \frac{[H^+][HA^-]}{[H_2 A]}$$ coming from $$H_2A \rightarrow H^+ + HA^-$$ The Bob (2004 ©)
P: 22,712

## titrate sulphuirc acid with sodium hydroxide

 Quote by GeneralChemTutor The pH of an equimolar sodium hydroxide and sulfuric acid is 7
It is not. Equimolar means 'Having an equal number of moles'. pH will be much lower, depending on the H+ concentration.

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P: 22,712
 Quote by The Bob I installed a system that was posted up here from the 'few questions' thread (called BATE) and it says that for an equal amount of $$0.01$$ $$mol$$ $$dm^{-3}$$ concentrations of each susbstance I should get an overall pH of 6.17. I am trying to find out how to get to this number and how I can explain it along the way.
The problem is, you will not get this number, because it is wrong - and I have no idea how did you get it using BATE :( Can you give some more details?

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BATE - pH calculations, titration curves, hydrolisis
HW Helper
P: 1,769
 It is not. Equimolar means 'Having an equal number of moles'. pH will be much lower, depending on the H+ concentration.
ok, first of all, as long as there is a strong acid/base component present all of the base/acid component will be neutralized (think, la chatelier). In all cases of titrations, assuming equimolar reaction, the final pH is actually pertains to the salt product obtained from the acid base reaction.

If you're referring to to the second ionization of sulfuric acid, it is negligible compared to the first. Look up the ionization constants yourself.
 Emeritus Sci Advisor PF Gold P: 11,154 The pH comes from the concentration of the acid/base as well as the dissociation constant (or you can use Ka/Kb). If the dissociation constants are very close to 1 (Ka/Kb large), the pH is roughly a function of concentration alone.
P: 22,712
 Quote by GeneralChemTutor In all cases of titrations, assuming equimolar reaction, the final pH is actually pertains to the salt product obtained from the acid base reaction.
pH of endpoint depends on the salt present at endpoint regardless of whether the reaction is equimolar or is not.

 If you're referring to to the second ionization of sulfuric acid, it is negligible compared to the first. Look up the ionization constants yourself.
I remember them so I don't have to check. pKa1=-3, pKa2 = 2. Difference is large but pKa2 is far from being negligible. NaHSO4 solutions are highly acidic - 0.01M solution has pH 2.21, not 7 as you stated.

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P: 1,116
 Quote by Borek The problem is, you will not get this number, because it is wrong - and I have no idea how did you get it using BATE :( Can you give some more details?
I tried it again and I didn't get the same answer. To be honest because I do not understand about the dissonance of the acids/bases I can't use the system properly.

 Quote by Borek I remember them so I don't have to check. pKa1=-3, pKa2 = 2. Difference is large but pKa2 is far from being negligible. NaHSO4 solutions are highly acidic - 0.01M solution has pH 2.21, not 7 as you stated.
This is what I want to know. How to work it out and how to apply it.

Cheers.

HW Helper
P: 1,769
 pH of endpoint depends on the salt present at endpoint regardless of whether the reaction is equimolar or is not.
If it is equimolar than it will depend on the salt alone, if not we'll need to account for the excess of the acid/base.

 NaHSO4 solutions are highly acidic - 0.01M solution has pH 2.21, not 7 as you stated.
I never said that a NaHS04 solution, an acidic solution, was neutral. The pH of HSO4- is not 2.21. The pH due to the first acid dissociation alone is 2.

 Difference is large but pKa2 is far from being negligible.
An acid/base classified as strong is deemed for the most part to dissociate almost to completion in water, in this case $K_a$ will be very large (Gokul...way above 1) since $K_a= \frac{[H^+][anion]}{[H_SO_4]}$. $K_{a2}$ is actually $1.2 x 10^{-2}$. There's really no comparison.

Bob, solve this equation for x, assuming a original solution of sulfuric acid of .01M and sodium hydroxide .01M...

$$.012= \frac{[x]^2}{[.01M-x]}$$

Subtract from pH 7. What you can also do is to actually calculate the pH of sulfuric acid as a whole, including Ka for both its acids. After finding the total concentration of H+ produced, find the excess of the latter by subtracting from the concentration of hydroxide. This would be more accurate.

$$.012= \frac{[.01+x][x]}{[.01-x]}$$, solve for x. $[H_3O^+]=[x+.01]$.

I'm sure you can do the rest.

You'll find that the pH of an equimolar solution of sodium hydroxide and sulfuric acid is roughtly 6.9, perhaps a bit higher.
P: 1,116
 Quote by GeneralChemTutor I'm sure you can do the rest.
Wouldn't be too sure about that. I am studying this one my own and this is the only place I can get help.

So how do you know that $$K _{a2}$$ is 0.012 and why is it that you are using $$K _{a2}$$???

P: 22,712
 Quote by GeneralChemTutor I never said that a NaHS04 solution, an acidic solution, was neutral. The pH of HSO4- is not 2.21. The pH due to the first acid dissociation alone is 2.
On 03-06-2005 at 11:27 you stated:

 The pH of an equimolar sodium hydroxide and sulfuric acid is 7
Now you deny it. Further discussion makes no sense.

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 P: 1,116 Here is the problem. I understand nothing at all to do with working out pHs or pOHs. I only found out yesterday that there was a unit pOH. The problem is I have 10 cm3 of Sodium Hydroxide and 10 cm3 of Sulphuric Acid. I mix them together and they, obviously, create an overall pH. I need to work out what that pH is. How do I do it, using the formulas that are needed and an explaination to what each bit is for and means, please. I would be really, really grateful if someone could explain it to be and, more or less, drag me through it. Thanks. The Bob (2004 ©)
P: 22,712
 Quote by The Bob This is what I want to know. How to work it out and how to apply it.

Dissociation equation:

$$AcH \leftrightarrow Ac^- + H^+$$

Equilibrium constant:

$$K_a = \frac{[Ac^-][H^+]}{[AcH]}$$

If you know total concentration - say it is $$C_a$$ - you may calculate $$H^+$$ concentration assuming - in accordance with reaction equation - that $$[Ac^-] = [H^+]$$. If so, undissociated acid concentration is $$[AcH] = C_a - [H^+]$$ and

$$K_a = \frac{[H^+]^2}{C_a - [H^+]}$$

which is a quadratic equation. Solve it for $$[H^+]$$:

$$[H^+] = \frac{-K_a + \sqrt{K_a^2 + 4 K_a C_a}}{2}$$

and you know how to perform pH calculation of the weak acid solution.

Assumption that $$[Ac^-] = [H^+]$$ is not always valid - to be precise you should take account of $$[H^+]$$ ions from the water autodissociation. For not very weak acids and not very diluted solutions assumption holds.

Now get back to the sulphuric acid and its equimolar solution with sodium hydroxide - which is the same as $$NaHSO_4$$ solution. You may treat $$HSO_4^-$$ the same way acetic acid was treated above - first dissociation step was already 'consumed' by neutralization and is not influencing the situation (that's not always the case - here is).

As I already wrote $$pK_{a2}$$ equals 2 - so $$K_{a2}$$ (which describes second dissociation equilibrium) equals 0.01. Put it into the equation for the $$[H^+]$$ together with the 0.01 concentration (in the real titration you will have to calculate the dilution factor) and you will get

$$[H^+] = \frac{-0.01 + \sqrt{0.01^2 + 4*0.01*0.01}}{2} = 0.00618$$

and pH = 2.21

Hope that helps.

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BATE - pH calculations, titration curves, hydrolisis
 Sci Advisor HW Helper P: 1,769 Borek, I mentioned a solution of the strong acid base, however you were implying that I had made a statement where an pure acid solution had a pH of 7. Bob, I'm not able to explain everything about acid/base chemistry right here, what you'll need to do is read up on your own. For now I'll explain a simple case...what is the pH of a solution of .01M sulfuric acid? We'll assume that the first acid dissociates to completion, this will give .01M of [H+] and [HSO4-] upon dissociation. The pH at this point can be calculated by taking the negative log of the concentration of [H+]....$pH=-log[H^+]$, at this point the pH is 2. However, if you wish to be absolutely exact the you'll need to incorporate the second ionization which has its own dissociation constant $K_{a2}$. This is where this equation comes in $$K_{a2}= \frac{[H+][SO_4^{2-}]}{[HSO_4^-]}$$ the initial concentration of [H+] is .01M, 0 for the sulfate, and .01 for the acid. Thus when x M of the acid has dissociated, $$.012= \frac{[.01+x][x]}{[.01-x]}$$ Solving for x and then finding [.01 + x] will give you the hydronium concentration of which the pH is approximately 1.9.
P: 1,116
 Quote by Borek Lets start with some easier, classic case - acetic acid (AcH). $$AcH \leftrightarrow Ac^- + H^+$$ Equilibrium constant: $$K_a = \frac{[Ac^-][H^+]}{[AcH]}$$ If you know total concentration - say it is $$C_a$$ - you may calculate $$H^+$$ concentration assuming - in accordance with reaction equation - that $$[Ac^-] = [H^+]$$
I can accept all of this. This is all fine.

 Quote by Borek If so, undissociated acid concentration is $$[AcH] = C_a - [H^+]$$ and $$K_a = \frac{[H^+]^2}{C_a - [H^+]}$$
This is where I get lost. I don't see what you have done and why.

The rest of this post I can, again, accept, so long as if I try another one I can run it through here.

 Quote by Borek and pH = 2.21
And yes, BATE says that too.

Cheers for the help so far.

 Quote by GeneralChemTutor $$.012= \frac{[.01+x][x]}{[.01-x]}$$